A 120 Ω resistor is in parallel with a capacitor having Xc = 40 Ω, both across a 12 V AC source. What is the magnitude of the total circuit impedance |Z|?

Introduction / Context:
For parallel RC circuits, it is often easier to work in admittance (Y) because conductances and susceptances add directly. The magnitude of impedance is the reciprocal of the magnitude of admittance.

Given Data / Assumptions:

• Resistive branch: R = 120 Ω ⇒ G = 1/R = 0.008333 S.
• Capacitive branch: Xc = 40 Ω ⇒ Bc = −1/Xc = −0.025 S (capacitive susceptance).
• Ideal sinusoidal conditions.

Concept / Approach:

Total admittance: Y = G + jBc. Magnitude |Y| = √(G^2 + Bc^2). Then |Z| = 1 / |Y|.

Step-by-Step Solution:

Verification / Alternative check:

Sanity check: Parallel combinations lower impedance below the smallest branch magnitude (min of 120 Ω and 40 Ω is 40 Ω). The result, ~37.9 Ω, is slightly below 40 Ω, which is consistent.

Why Other Options Are Wrong:

3.7 Ω is an order-of-magnitude error. 14,400 Ω and 4,800 Ω result from misapplied series reasoning or arithmetic inversions.

Common Pitfalls:

Adding impedances directly in parallel; forgetting to invert admittance to return to impedance.

Final Answer:

37.9 Ω